JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    Direction: A voltage source \[V={{V}_{0}}\,\sin \,(100t)\] is connected to a black box in which there can he either one element out of L, C, R or any two of them connected in series.
    At steady state the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown.
    If AC source is removed, the circuit is shorted and then at \[t=0\], a battery of constant \[emf\] is connected across the black box. The current in the circuit will

    A)  increase exponentially with constant \[=4\times {{10}^{-3}}s\]

    B)  decrease exponentially with time constant \[=1\times {{10}^{-2}}s\]

    C)  oscillate with angular frequency \[20\,{{s}^{-1}}\]

    D)  first increase and then decrease

    Correct Answer: B

    Solution :

     \[\Delta \theta =\omega \Delta t=\frac{\pi }{4}\] \[\tan \,\theta =\frac{X}{R}\] \[\Rightarrow \]            \[X=R\] Since, current leads the voltage the circuit consists of R and C. and      \[{{i}_{0}}=\frac{{{V}_{0}}}{z}\] \[\therefore \]    \[z=\frac{{{V}_{0}}}{{{i}_{0}}}=\frac{100}{\sqrt{2}}=50\sqrt{2}\] Now,    \[R\sqrt{2}=50\sqrt{2}\] \[\Rightarrow \]            \[R={{X}_{C}}=50\] \[{{X}_{C}}=\frac{1}{C\omega }=50\] \[\Rightarrow \]            \[C=\frac{1}{50\,\omega }=200\,\mu F\] \[\tau =RC=50\times 200\times {{10}^{-6}}=1\times {{10}^{-2}}s\]

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